What is the extreme f(x,y)=x^2y-2xy+y^2+3 ?

The extreme f(x,y)=x^2y-2xy+y^2+3 is Saddle(2,0),Saddle(0,0),Minimum(1, 1/2)

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extreme f(x,y)=x^2y-2xy+y^2+3

extreme points

f (x, y) = x^{2} y - 2 x y + y^{2} + 3

Saddle (2, 0), Saddle (0, 0), Minimum (1, \frac{1}{2})

Solution steps

Find the critical points:

(2, 0), (0, 0), (1, \frac{1}{2})

\frac{\partial ^{2} f}{\partial x ^{2}} = 2

D (x, y) = 4 y - 4 x^{2} + 8 x - 4

Check the sign of at each critical point

Check critical point

(2, 0) :

Saddle

Check critical point

(0, 0) :

Saddle

Check critical point

(1, \frac{1}{2}) :

Minimum

Saddle (2, 0), Saddle (0, 0), Minimum (1, \frac{1}{2})

x^{2} y + y^{3} - 12 y

f (x, y) = 2 x^{2} + 3 x y + y^{2} + a x + 5

f (x, y) = (x - y) (1 - x y)

f (x) = 2 x^{2} - 4 x - 1

f (x) = 3 x^{4} + 16 x^{3} + 24 x^{2} + 32

What is the extreme f(x,y)=x^2y-2xy+y^2+3 ?
The extreme f(x,y)=x^2y-2xy+y^2+3 is Saddle(2,0),Saddle(0,0),Minimum(1, 1/2)